Edexcel A Level Chemistry

Revision Notes

6.3.2 Chromium

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Reduction & Oxidation of Chromium Species

  • For chromium we need to consider the following standard electrode potential values
  • We will use zinc and hydrogen peroxide as oxidising agents
  • The half equations are arranged from high negative EΘ at the top to high positive EΘ at the bottom
    • The best reducing agent is the top right species (Zn (s))
    • The best oxidising agent is the bottom left species (Cr2O72-(aq))

reduction-and-oxidation-of-chromium-species-table

Oxidation from +3 to +6

  • The two half equations we need to consider are 3 and 4
  • Chromiums oxidation number changes from +6 to +3 in half equation 3
  • The EΘ value for half equation 3 is more negative than the EΘ for half equation 4
    • Cr(OH)3 (aq) is the best reducing agent
    • H2O2 (aq) is the best oxidising agent
  • We can obtain the overall equation by reversing half equation 3 and combining it with equation 4
    • When adding half equations remember to multiply them so each has the same number of electrons

 2Cr(OH)(aq) + 4OH- (aq) + 3H2O2 (aq) → 2CrO42- (aq) + 8H2O (l)

  • This reaction is carried out in alkaline conditions due to the presence of OH- ions in the equation

Reduction from +6 to +3

  • The two half equations we need to consider are 1 and 5
  • Chromiums oxidation number changes from +6 to +3 in half equation 3
  • The EΘ value for half equation 1 is more negative than the EΘ for half equation 5
    • Zn is the best reducing agent
    • Cr2O72- is the best oxidising agent
  • We can obtain the overall equation by reversing half equation 1 and combining it with equation 5
    • When adding half equations remember to multiply them so each has the same number of electrons

 Cr2O72- (aq) + 14H+ (aq) + 3Zn (s) → 2Cr3+ (aq) + 7H2O (l) + 3Zn2+ (aq)

  • This reaction is carried out under acidic conditions due to presence of H+ in the equation

Reduction from +3 to +2

  • The Cr3+ ion can be further reduced by zinc
  • The two half equations we need to consider are 1 and 2
  • Chromiums oxidation number changes from +3 to +2 in half equation 3
  • The EΘ value for half equation 1 is more negative than the EΘ for half equation 2
    • Zn (s) is the best reducing agent
    • Cr3+ (aq) is the best oxidising agent
  • We can obtain the overall equation by reversing half equation 1 and combining it with equation 2
    • When adding half equations remember to multiply them so each has the same number of electrons

 2Cr3+ (aq) + Zn  (s) → 2Cr2+ (aq) + Zn2+ (aq)

  • As this reaction is a further step from the previous reduction this reaction is also carried out under acidic conditions

The Dichromate(VI) - Chromate(VI) Equilibrium

  • The chromate CrO42- and dichromate Cr2O72- ions can be converted from one to the other by the following equilibrium reaction

2CrO42- (aq) + 2H(aq) ⇌ Cr2O72- (aq) + H2O (l) 

  • Chromate(VI) ions are stable in alkaline solution, but in acidic conditions the dichromate(VI) ion is more stable
  • Addition of acid will push the equilibrium to the dichromate
    • This results in a colour change from yellow to orange
  • Addition of alkali will remove the H+ ions and push the equilibrium to the chromate
  • This is not a redox reaction as both the chromate and dichromate ions have an oxidation number of +6
    • This is an acid base reaction

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